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The administration at a college wishes to estimate, to within two percentage points, the proportion of all its entering freshmen who graduate within four years, with \(90 \%\) confidence. Estimate the minimum size sample required.

Short Answer

Expert verified
The minimum sample size required is 1688.

Step by step solution

01

Understanding the Problem

We need to determine the minimum sample size required to estimate the proportion of entering freshmen who graduate within four years, with a confidence level of 90% and a margin of error of 2 percentage points.
02

Identifying Key Components

For this exercise, the key components include the desired confidence level (90%), the margin of error (2 percentage points or 0.02 when expressed as a decimal), and the worst-case proportion estimate (usually assumed to be 0.5 for unknown proportions).
03

Using the Sample Size Formula

The formula for sample size when estimating a proportion is: \[ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) \]where \( n \) is the sample size, \( Z \) is the Z-score for the desired confidence level, \( p \) is the estimated proportion, and \( E \) is the margin of error.
04

Calculating the Z-score

For a confidence level of 90%, we find the Z-score from a standard normal distribution table. The Z-score for 90% confidence is approximately 1.645.
05

Substitute Values into the Formula

Using \( Z = 1.645 \), \( p = 0.5 \) (assuming worst-case scenario), and \( E = 0.02 \), we can substitute into the formula:\[ n = \left( \frac{1.645^2 \cdot 0.5 \cdot (1-0.5)}{0.02^2} \right) \]
06

Solving the Formula

Calculate:\[ 1.645^2 = 2.70 \]Now substitute into the expression:\[ n = \frac{2.70 \cdot 0.5 \cdot 0.5}{0.0004} \]\[ n = \frac{0.675}{0.0004} = 1687.5 \]Since we cannot have a fraction of a sample size, we round up to the next whole number.
07

Concluding the Minimum Sample Size

After rounding 1687.5 up, we get a sample size of at least 1688. Hence, the minimum sample size required is 1688.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Level
The concept of a confidence level is crucial in statistics, as it tells us how certain we are that our sample reflects the true population parameter. Think of it like the reliability of your predictions. When you have a 90% confidence level, you are saying that 90 out of 100 times, the sample result will fall within the desired margin of error of the true population value.
This level of confidence is associated with a Z-score, which is a standardized score that corresponds to the desired confidence. For example, a 90% confidence level implies a Z-score of approximately 1.645.
Why Z-scores? They come from the standard normal distribution, which is a way of standardizing datasets to allow comparisons.
  • Confidence Level tells how sure we are about our results.
  • It is usually given as a percentage (e.g., 90%, 95%).
  • Z-score is used to calculate this level and determines the margin of error.
Understanding this concept helps ensure that your data is sufficiently reliable for making predictions.
Margin of Error
The margin of error is a statistic expressing the amount of random sampling error in a survey's results. It provides a range within which the true population parameter is expected to fall.
For this specific problem, the margin of error is set at 2 percentage points or 0.02 as a decimal. This means that we want our estimation to be within 2% of the true proportion of freshmen who graduate within four years.
The formula to calculate the required sample size uses the margin of error as one of its key components.
  • Margin of Error defines the level of precision of the results.
  • It is directly related to the confidence level.
  • Smaller margin of error demands a larger sample size.
Always remember, a smaller margin of error increases the preciseness of your results but it often requires more data.
Proportion Estimation
Proportion estimation is a statistical technique used to calculate the percentage of a population having a specific characteristic. In sample size calculations, especially when the actual proportion isn't known, it's common to assume a worst-case scenario estimation of 0.5 or 50%. This is because it maximizes the product of the proportion \( p \) and \( (1-p) \), ensuring the most conservative sample size.
In our particular exercise, the administration wants to estimate the proportion of freshmen graduating in four years.
  • When the proportion estimate \( p \) is unknown, use 0.5 for conservative calculations.
  • The estimation helps define necessary data collection strategies.
  • It's integral in ensuring decisions are data-driven and accurate.
By understanding proportion estimation, you are equipped to conduct effective sampling and make informed conclusions about your population of interest.

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Most popular questions from this chapter

A random sample of 185 college soccer players who had suffered injuries that resulted in loss of playing time was made with the results shown in the table. Injuries are classified according to severity of the injury and the condition under which it was sustained. $$ \begin{array}{|c|c|c|c|} \hline & \text { Minor } & \text { Moderate } & \text { Serious } \\ \hline \text { Practice } & 48 & 20 & 6 \\ \hline \text { Game } & 62 & 32 & 17 \\ \hline \end{array} $$ a. Give a point estimate for the proportion \(p\) of all injuries to college soccer players that are sustained in practice. b. Construct a \(95 \%\) confidence interval for the proportion \(p\) of all injuries to college soccer players that are sustained in practice. c. Give a point estimate for the proportion \(p\) of all injuries to college soccer players that are either moderate or serious.

A random sample is drawn from a population of unknown standard deviation. Construct a \(99 \%\) confidence interval for the population mean based on the information given. a. \(\quad n=49, x-17.1, s=2.1\) b. \(\quad n=169, x-=17.1, s=2.1\)

The body mass index (BMI) was measured in 1,200 randomly selected adults, with the results shown in the table. a. Give a point estimate for the proportion of all men whose BMI is over 25 . b. Assuming the sample is sufficiently large, construct a \(99 \%\) confidence interval for the proportion of all men whose BMI is over 25 c. Give a point estimate for the proportion of all adults, regardless of gender, whose BMI is over \(25 .\) d. Assuming the sample is sufficiently large, construct a \(99 \%\) confidence interval for the proportion of all adults, regardless of gender, whose BMI is over \(25 .\)

A government agency was charged by the legislature with estimating the length of time it takes citizens to fill out various forms. Two hundred randomly selected adults were timed as they filled out a particular form. The times required had mean 12.8 minutes with standard deviation 1.7 minutes. Construct a \(90 \%\) confidence interval for the mean time taken for all adults to fill out this form.

Wild life researchers trapped and measured six adult male collared lemmings. The data (in millimeters) are: \(104,99,112,115,96,109 .\) Assume the lengths of all lemmings are normally distributed. a. Construct a \(90 \%\) confidence interval for the mean length of all adult male collared lemmings using these data. b. Convert the six lengths in millimeters to lengths in inches using the conversion \(1 \mathrm{~mm}=0.039\) in (so the first datum changes to \((104)(0.039)=4.06\) ) . Use the converted data to construct a \(90 \%\) confidence interval for the mean length of all adult male collared lemmings expressed in inches. c. Convert your answer in part (a) into inches directly and compare it to your answer in (b). This illustrates that if you construct a confidence interval in one system of units you can convert it directly into another system of units without having to convert all the data to the new units.

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